# Reversible cycle and adiabatic expansion

1. Sep 21, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
One mole of a monatomic ideal gas is taken through the reversible cycle shown below. Process bc is an adiabatic expansion, with Pb = 10.0 atm and Vb = 1.00 x 10-3 m3. Find (a) the energy added to the gas as heat (b) the energy leaving the gas as heat, (c) the net work done by the gas, and (d) the efficiency of the cycle.

I attached the diagram of the cycle:

2. Relevant equations

Ideal gas law:
pv=nRT

dQ=dW + dU

3. The attempt at a solution

For the first part I need to find the Q into the system which would be from dQ from a to b.

so dQ= dU + dW
dW=0 since there is constant volume. Meaning,
dQ=dU

Since the gas is a monotomic ideal gas, that means
dU= (3/2)nR(Tb-Ta), since n=1 that means:
dU=(3/2)R(Tb-Ta) and thus
dQ=(3/2)R(Tb-Ta)

However, I dont know how to find out the temperatures. I tried figuring them out using the adiabatic equations assuming gamma=1.67 (for an ideal gas)

where PV^gamma=constant
and V^(gamma-1)T= constant
but I'm getting values like 15 K, and I know that is unrealistic, especially for a gas. Any help/tips would be appreciated!!!

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